jesusricardobas
jesusricardobas jesusricardobas
  • 17-11-2019
  • Mathematics
contestada

The scores of the athletes are 32,45,33,22,50,48. Find the variance.

Respuesta :

beritop1089
beritop1089 beritop1089
  • 22-11-2019

Answer:

Step-by-step explanation:

First, find the mean ; (32+45+33+22+50+48) / 6 = 230/6

Mean = 38.33

Variance =[tex]\frac{SUM(X-meanX)^{2} }{n-1}[/tex]

n = sample size

Variance = [tex]\frac{(32-38.33)^{2} +(45-38.33)^{2}+(33-38.33)^{2} +(22-38.33)^{2}+(50-38.33)^{2}+(48-38.33)^{2} }{5}[/tex]

Variance = [tex]\frac{(44.4889 +40.0689 +28.4089 + 266.6689 +136.1889 +93.5089)}{5}[/tex]

= [tex]\frac{609.3334}{5}[/tex]

= 121.8667

Therefore, variance = 121.87

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